Mixture problems 2 ways!

You need a 15% acid solution for a certain test, but your supplier only ships

a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the

supplier make a 15% solution, you decide to mix 10% solution with 30%solution, to make your own 15% solution. You need 10 liters of

the 15% acid solution. How many liters of 10% solution and 30% solution should you

use?

You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution

with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?

Mixture problems are CA CA!CA+CA=CA or, more specifically, C1A1+C2A2=CTAT

C= Concentration or Cost depending on the problemA=AmountThe subscripts are the 1st item, the 2nd item, and the total (or mixture)

The following graphic simply gets all of the numbers organized in order to put them into the formula.

C A

1 10 x

2 30 10-x

T 15 10

The amounts of 1 & 2 must add to make the total amount! In this problem, we know the total (10 liters) So, we make #1 = x, then #2 must be 10-x so that 1 & 2 together add to 10.

Now, we put the numbers into the formula C1A1+C2A2=CTAT

10x + 30(10-x) = 15•10 10x + 300-30x = 150 -20x + 300 = 150 -300 -300 -20x = -150 -20 -20 X= 7.5 (this is the 10% solution)

10-x10-7.52.5(this is the 30% solution)

• Same problem, different way:

You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution

with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?

1 2mix

10 30C

A x 10-x10 L

15

15-10 = 5

30-15=15

5x = 15(10-x)5x = 150 – 15x

+15x +15x20x = 15020 20 x = 7.5 LOf the 10% solution

10-x10-7.52.5Of the 30% solution

This graphic also organizes the information, but goes one step further and takes care of some of the arithmetic too.

How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to

produce a mixture worth $1.00 a pound?

C A

1 $1.20 x

2 $0.90 10

T $1.00 x + 10

The amounts of 1 & 2 must add to make the total amount! In this problem, we know the amount of #2 is 10 pounds, So, we make #1 = x, then the total must be x +10.

Now, we put the numbers into the formula C1A1+C2A2=CTAT

$1.20x + .90•10 = $1.00(x+10) 1.20x + 9 = 1.00x + 10 -1.00x -1.00x .20x +9 = 10 -9 -9 .20x = 1 .20 .20 X= 5 pounds of the $1.20 chocolate

How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to

produce a mixture worth $1.00 a pound?

1 2mix

$1.20 $0.90C

A x 1010 + x

$1.00

$1.20-1.00=.20

$1.00-.90=.10

.20x = .10 • 10

.20x = 1

.20 .20

Of the $1.20 chocolate

x = 5